Seven consecutive positive integers have a sum of 91. What is the largest of these integers?
Answer: We can let the largest of the integers be $x$. Then the other six integers are $x-1$, $x-2$, $x-3$, $x-4$, $x-5$, and $x-6$. Then, finding the sum of these integers, we have \begin{align*}x+(x-1)+(x-2)+(x-3)+(x-4)+(x-5)+(x-6)&=91 \\ 7x-21&=91 \\ 7x&=112 \\ {x}&=16\end{align*} Therefore the largest of the integers is $\boxed{16}$.